A box is given a push so that it slides across the floor. How far will it go, given the coefficient of kinetic friction is 0.20 and the push imparts an initial speel of 4.0 meter/second.
2006-09-27 21:09:07 · 4 answers · asked by Pat M 1 in Science & Mathematics ➔ Physics
You can use the Work-Energy Theorem:
W = AK= Kf-Ki= 0 - 1/2mv^2=-umgd, W <0, because the work is produced by friction force.
then d = v^2 / 2ug = 4.081m
2006-09-27 21:23:24 · answer #1 · answered by Juan D 3 · 0⤊ 0⤋
4 m
2006-09-28 04:53:13 · answer #2 · answered by amir sabet sarvestani 1 · 0⤊ 2⤋
you have a kinetic energy due to the velocity of 4 m/s that a force f (friction) will "cosnume" till the object reaches a stop.
therefore
1/2 m * U init ^2 = F * d, but
F= n * N = 0.2 * m * g ==>>
d= U^2 / 2 * 0.2 * g = > d = 4.077m
2006-09-28 04:47:50 · answer #3 · answered by Emmanuel P 3 · 0⤊ 0⤋
It will go until its' kinetic energy is zero.
Ke(x) = mv²/2 - 2*(.2m*x)*(x) = 0
Doug
2006-09-28 04:18:39 · answer #4 · answered by doug_donaghue 7 · 1⤊ 0⤋